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# Let $f(x) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1$ , $\left| x \right| > 1$ , if $\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f(x)} \right)} \right)$ and $y\left( {\sqrt 3 } \right) = \dfrac{\pi }{6}$ , then $y\left( { - \sqrt 3 } \right)$ is equal to :(A) $- \dfrac{\pi }{6}$(B) $\dfrac{{5\pi }}{6}$(C) $\dfrac{{2\pi }}{3}$(D) $\dfrac{\pi }{3}$

Last updated date: 03rd Aug 2024
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Hint: Since in the given question we need to determine the value of $y\left( { - \sqrt 3 } \right)$ so that means we need to find the value of $y(x)$ , in the first part we will be determine the value of $y(x)$ by integrating $\dfrac{{dy}}{{dx}}$ and it can seen that $y(x)$ involves the function $f(x)$ that means we need to simplify it get the value that can be substituted to get the required value of $y(x)$.

Given
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f(x)} \right)} \right)$
Integrating both side
$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{1}{2}} } \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f(x)} \right)} \right)$
Integrating of any differential function $\dfrac{d}{{dx}}\left( {g(x)} \right) = g(x) + c$
So by using this
$y = \dfrac{1}{2}{\sin ^{ - 1}}(f(x)) + c$
Now we know that $f(x) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1$
Now we made some modification in given above equation
Let assume ${\tan ^{ - 1}}x = \theta$ ( this assumption we made just to making easy to this equation)
And we know that
${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$
From this we can written as ${\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$ and we assume ${\tan ^{ - 1}}x = \theta$
So by putting this thing in $f(x) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1$
$f(x) = {\left( {\sin \theta + \sin \left( {\dfrac{\pi }{2} - \theta } \right)} \right)^2} - 1$
Using
$\because {\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x = \dfrac{\pi }{2} - \theta$
Now, we know that $\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta$
So, $f(x) = {\left( {\sin \theta + \cos \theta } \right)^2} - 1$
Now, we know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
${\left( {\sin \theta + \cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta$
Now $\because {\sin ^2}\theta + {\cos ^2}\theta = 1$
So from here we get
$\Rightarrow$$f(x) = 1 + 2\sin \theta \cos \theta - 1 Simplifying the above, we get \Rightarrow$$f(x) = 2\sin \theta \cos \theta$
Using
$\because 2\sin \theta \cos \theta = \sin 2\theta$
we get
$\therefore f(x) = \sin 2\theta$
Now, put this value in
$y = \dfrac{1}{2}{\sin ^{ - 1}}(f(x)) + c$
we get
$\Rightarrow$$y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin 2\theta } \right) + c Now replace value of \theta by {\tan ^{ - 1}}x So, \Rightarrow$$y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right) + c$
Now given that $y\left( {\sqrt 3 } \right) = \dfrac{\pi }{6}$
Hence,
$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}\sqrt 3 )} \right) + c we get \because {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3} By putting this value we get \Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \dfrac{{2\pi }}{3}} \right) + c$
Simplifying the above
$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right) + c We get, \Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2} \times \dfrac{\pi }{3} + c$
From here
$c = 0$
Now our $y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right)$
Now we have to find $y\left( { - \sqrt 3 } \right)$
So put $x = - \sqrt 3$ in $y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right)$
We get
$\Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \left( {(2{{\tan }^{ - 1}}( - \sqrt 3 )} \right)} \right) using \because {\tan ^{ - 1}}( - \sqrt 3 ) = - \dfrac{\pi }{3} By putting this we get \Rightarrow y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \left( { - \dfrac{{2\pi }}{3}} \right)} \right) using \because \sin \left( { - \dfrac{{2\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2} By putting this we get \Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right)$
using
$\because {\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = - \dfrac{\pi }{3}$
By putting this we get
$\Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( { - \dfrac{\pi }{3}} \right) So it become \Rightarrow$$y( - \sqrt 3 ) = - \dfrac{\pi }{6}$

So option A is the correct answer.

Note: The best way to determine the specific value involving trigonometric function is to consider or to convert into the short term as in this the function $f(x)$ is simplified to short term that makes the solution extremely easy to solve otherwise it will be quite difficult to determine the result and if we need to determine the value of constant in that case we use the initial condition as in the above question.